König's Infinity Lemma and Aronszajn Trees

Question

I am working through the notes of my Set Theory lecture.
There my professor wrote:

'Is there an uncountable $\kappa$ such that König's Infinity Lemma holds for $\kappa$?

There are models where $\aleph_2$ is such. But, if CH holds then there are $\aleph_2$-Aronszajn trees...'

What could he mean with the three dots? What would follow if we had $\aleph_2$ Aronszajn trees?

(König's Infinity Lemma, as we had it in the lecture:
If $T$ is a tree of hight $\omega$ with all levels finite, then $T$ has an infinite branch.

Aronszajn tree: hight $\omega_1$, levels at most countable and no uncountable branch.)

Answer 1

The $\kappa$ version of König's lemma states that if a tree $T$ has height $\kappa$ and each level has size strictly less than $\kappa$, then $T$ has a branch of length $\kappa$.

A $\kappa$ Aronszajn tree is a tree of height $\kappa$ and levels of size strictly less than $\kappa$ which has no branches of length $\kappa$.

That is, precisely by definition, a $\kappa$ Aronszajn tree is a counterexample to the $\kappa$ version of König's lemma.

The question is asking whether we can prove (in $\mathsf{ZFC}$, I assume) that there is a $\kappa$ for which the $\kappa$ König's lemma holds. The consistency of the existence of $\aleph_2$ Aronszajn trees shows that we cannot take $\kappa=\aleph_2$ in general.

The tree property at $\kappa$ is the statement that there are no $\kappa$ Aronszajn trees (that is, the $\kappa$ König's lemma holds). If $\kappa$ is inaccessible, having the tree property is precisely equivalent to $\kappa$ being weakly compact. At other cardinals, the tree property is more delicate, and is connected with the existence of large cardinals in inner models. More precisely: Mitchell proved that if there is a weakly compact cardinal, then there is a forcing extension where $\aleph_2$ has the tree property. We can replace $\aleph_2$ here with the successor of any uncountable regular cardinal without difficulties. Conversely, Silver observed that if an uncountable cardinal $\kappa$ has the tree property, then $\kappa$ is weakly compact in $L$.

The difficulties increase if we ask for a cardinal and its successor both to have the tree property. The strength of this has been investigated by several people, notably Foreman-Magidor-Schindler, in The consistency strength of successive cardinals with the tree property, Journal of Symbolic Logic 66 (2001), pp. 1837–1847.

For the latest results on the problem of obtaining models with successive cardinals with the tree property, see this preprint by Neeman.

The upper bound needed by Neeman (many supercompacts) is huge compared with the lower bound in the three author paper (Projective determinacy). There is still plenty of work to do here.

You may find this question useful. Also, see Kanamori's book The higher infinite for references and proofs on this topic. Of note: The tree property at $\kappa$ implies that $\kappa$ is regular, and if it is the successor of $\lambda$, then either $\lambda$ is singular, or else $2^{<\lambda}>\lambda$, this was first observed by Specker (the case $\lambda=\omega_1$ is the $\mathsf{CH}$ result stated in the question).

Additional results can be found in the references listed in the papers mentioned above. A result absent there is the one mentioned by Asaf in comments: Arthur Apter has recently showed that it is consistent, relative to the existence of a proper class of supercompact cardinals, that $\mathsf{ZF}+\mathsf{DC} +$ Every successor cardinal is regular, every limit cardinal is singular, and every successor cardinal has the tree property. See A remark on the tree property in a choiceless context, Archive for mathematical logic, 50, (2011), 585-590. Whether something like this is possible with full choice is currently open, Neeman's result mentioned above is the state of the art.

Answer 2

Recall that an $\aleph_2$-Aronszajn tree is a tree of height $\omega_2$ that every level of it is of size at most $\aleph_1$, but there is no path of length $\omega_2$.

This is exactly a counterexample to the Koenig's lemma for $\aleph_2$.